3

In the context of compound Poisson distribution we have:

  • $X \sim F$ a random variable;
  • $\{X_n\}_{n\in \mathbb{N}}$ be a i.i.d. sequence of copies of $X$ - i.e. $X_j \sim F\,\, \forall \, j \in \mathbb{N}$ and $X_i, X_j$ independent for all $i,j \in \mathbb{N}$.
  • $N\sim Po(\lambda)$ a Poisson random variable such that $N$ and $X_j$ are independent for all $j$.

Define the Poisson Sum as: $$Y = \sum_{j=1}^N X_{j}$$

Accordingly to this (Properties section), when we calculate $E(Y)$ or $V(Y)$, I would like to know at what point it is used that $X_j$ and $N$ are independent?

I do not know the importance and use of the independence! Help!

Jean-Claude Arbaut
  • 23,601
  • 7
  • 53
  • 88
PSE
  • 544

1 Answers1

3

I'm assuming

  1. $\{X_j\}_{j\in \mathbb{N}}$ is an iid sequence, i.e. they are identically distributed and mutually--not pairwise as you have it-- independent.
  2. $N$ is independent of $\{X_j\}_{j\in \mathbb{N}}$.

By Adam's Law, the mean of $Y$ is

$$\small \begin{aligned} E\left[\sum_{j=1}^N X_j \right] &=E\left[E\left[\sum_{j=1}^N X_j\Bigg|N \right]\right]\\ &=E\left[\sum_{j=1}^NE\left[ X_j\big|N \right]\right]&&\qquad (\text{linearity of }E[\cdot ])\\ &=E\left[\sum_{j=1}^NE\left[ X_j \right]\right]&&\qquad (N,X_j\text{ indep.})\\ &=E\left[NE\left[ X_1 \right]\right]&&\qquad (\{X_j\}_{j\in \mathbb{N}}\text{ identically distrib.})\\ &=E\left[N\right]E\left[ X_1 \right],&&\qquad (\text{linearity of }E[\cdot ])\end{aligned}$$

which is a special case of Wald's equation.

By Eve's law, the variance of $Y$ is

$$ \small\begin{aligned}\text{Var}\left[\sum_{j=1}^N X_j \right]&=E\left[\text{Var}\left[\sum_{j=1}^N X_j\Bigg|N \right]\right]+\text{Var}\left[E\left[\sum_{j=1}^N X_j\Bigg|N \right]\right]\\ &=E\left[\text{Var}\left[\sum_{j=1}^N X_j\Bigg|N \right]\right]+\text{Var}\left[NE[X_1]\right]&&(\text{see above})\\ &=E\left[\text{Var}\left[\sum_{j=1}^N X_j \right]\right]+\text{Var}\left[NE[X_1]\right]&&(N,\{X_j\}_{j\in \mathbb{N}} \text{ indep.})\\ &=E\left[\sum_{j=1}^N\text{Var}\left[ X_j \right]\right]+\text{Var}\left[NE[X_1]\right]&& (\{X_j\}_{j\in \mathbb{N}} \text{ indep.})\\ &=E\left[N\text{Var}\left[ X_1 \right]\right]+\text{Var}\left[NE[X_1]\right]&& (\{X_j\}_{j\in \mathbb{N}} \text{ identically distrib.})\\ &=E\left[N\right]\text{Var}\left[ X_1 \right]+\text{Var}\left[N\right](E[X_1])^2&& (\text{linearity of } E[\cdot ])\\ &= E\left[N\right]E\left[ X_1^2 \right]+\left(\text{Var}\left[N\right]-E\left[N\right]\right)(E[X_1])^2. \end{aligned}$$

For $N$ Poisson with rate parameter $\lambda$, we have $E[N]=\text{Var}[N]=\lambda$, so the above simplifies to

$$E[Y]=\lambda E[X_1]\\ \text{Var}[Y]=\lambda E\left[ X_1^2 \right],$$

as obtained in the wiki link.

Note the variance computation required stronger assumptions than the mean computation, which is typically the case to get a nice variance expression.

Golden_Ratio
  • 12,834
  • 1
    Further note in case you are worried about an infinite collection of RVs: https://math.stackexchange.com/questions/1537542/joint-distribution-of-an-infinite-collection-of-random-variables – Golden_Ratio Jan 22 '22 at 05:08
  • Related to the $V(Y)$ calculation, in the line after (see above),I don't understand why you directly removed the conditioning on N (on the first term sum in the third line) . I think that if $X_1,...,X_N$ are independent, then the variance is linear. So, the conditional on N are also linear. So: $V(X_1 +...+ X_N|N) = V(X_1|N)+ ... + V(X_N|N)$. I know that $E(X_j|N)= E(X_j)$, but i dont know if it remais for the variance: $V(X_j|N)=V(X_j)$. If it is true, this is the pass from the second line to fourth line. Am I wrong? – PSE Jan 22 '22 at 19:12
  • 1
    @PSE $V(X_1+...+X_N|N)=V(X_1|N)+...+V(X_N|N)$ would hold if the $X_j$s are conditionally independent (conditional on $N$). And independence does not generally imply conditional independence. Hence my statement of assumption 2, namely that $N$ is independent of the entire sequence of $X_j$s. – Golden_Ratio Jan 22 '22 at 19:19
  • ^which is stronger than the assumption I used to obtain the mean, which only used pairwise independence between $N$ and each $X_j$ – Golden_Ratio Jan 22 '22 at 19:21
  • What is conditionally independence? Is the same that mean-independence: $X$ is mean independent of $Y$ if $E(X|Y)=E(X)$? If yes, I thin that the independence implies mean-independence, acording this: https://statisticaloddsandends.wordpress.com/2021/12/11/what-is-mean-independence/ – PSE Jan 22 '22 at 19:33
  • 1
    @PSE Conditional independence is not mean independence. ; pls see https://en.wikipedia.org/wiki/Conditional_independence – Golden_Ratio Jan 22 '22 at 19:35
  • Thanks for clarifying the concepts. But I still don't understand why you removed the $N$ conditioning from the second to the third line pass? Even assuming that $N$ is independent of the entire sequence of $X's$, I don't understand :( – PSE Jan 22 '22 at 19:42
  • Let us continue this discussion in chat. – Golden_Ratio Jan 22 '22 at 19:43